Integrand size = 16, antiderivative size = 121 \[ \int \sqrt {x} (a-b x)^{5/2} \, dx=-\frac {5 a^3 \sqrt {x} \sqrt {a-b x}}{64 b}+\frac {5}{32} a^2 x^{3/2} \sqrt {a-b x}+\frac {5}{24} a x^{3/2} (a-b x)^{3/2}+\frac {1}{4} x^{3/2} (a-b x)^{5/2}+\frac {5 a^4 \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a-b x}}\right )}{64 b^{3/2}} \]
5/24*a*x^(3/2)*(-b*x+a)^(3/2)+1/4*x^(3/2)*(-b*x+a)^(5/2)+5/64*a^4*arctan(b ^(1/2)*x^(1/2)/(-b*x+a)^(1/2))/b^(3/2)+5/32*a^2*x^(3/2)*(-b*x+a)^(1/2)-5/6 4*a^3*x^(1/2)*(-b*x+a)^(1/2)/b
Time = 0.34 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.80 \[ \int \sqrt {x} (a-b x)^{5/2} \, dx=\frac {\sqrt {x} \sqrt {a-b x} \left (-15 a^3+118 a^2 b x-136 a b^2 x^2+48 b^3 x^3\right )}{192 b}+\frac {5 a^4 \arctan \left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a-b x}}\right )}{32 b^{3/2}} \]
(Sqrt[x]*Sqrt[a - b*x]*(-15*a^3 + 118*a^2*b*x - 136*a*b^2*x^2 + 48*b^3*x^3 ))/(192*b) + (5*a^4*ArcTan[(Sqrt[b]*Sqrt[x])/(-Sqrt[a] + Sqrt[a - b*x])])/ (32*b^(3/2))
Time = 0.19 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {60, 60, 60, 60, 65, 216}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \sqrt {x} (a-b x)^{5/2} \, dx\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {5}{8} a \int \sqrt {x} (a-b x)^{3/2}dx+\frac {1}{4} x^{3/2} (a-b x)^{5/2}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {5}{8} a \left (\frac {1}{2} a \int \sqrt {x} \sqrt {a-b x}dx+\frac {1}{3} x^{3/2} (a-b x)^{3/2}\right )+\frac {1}{4} x^{3/2} (a-b x)^{5/2}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {5}{8} a \left (\frac {1}{2} a \left (\frac {1}{4} a \int \frac {\sqrt {x}}{\sqrt {a-b x}}dx+\frac {1}{2} x^{3/2} \sqrt {a-b x}\right )+\frac {1}{3} x^{3/2} (a-b x)^{3/2}\right )+\frac {1}{4} x^{3/2} (a-b x)^{5/2}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {5}{8} a \left (\frac {1}{2} a \left (\frac {1}{4} a \left (\frac {a \int \frac {1}{\sqrt {x} \sqrt {a-b x}}dx}{2 b}-\frac {\sqrt {x} \sqrt {a-b x}}{b}\right )+\frac {1}{2} x^{3/2} \sqrt {a-b x}\right )+\frac {1}{3} x^{3/2} (a-b x)^{3/2}\right )+\frac {1}{4} x^{3/2} (a-b x)^{5/2}\) |
\(\Big \downarrow \) 65 |
\(\displaystyle \frac {5}{8} a \left (\frac {1}{2} a \left (\frac {1}{4} a \left (\frac {a \int \frac {1}{\frac {b x}{a-b x}+1}d\frac {\sqrt {x}}{\sqrt {a-b x}}}{b}-\frac {\sqrt {x} \sqrt {a-b x}}{b}\right )+\frac {1}{2} x^{3/2} \sqrt {a-b x}\right )+\frac {1}{3} x^{3/2} (a-b x)^{3/2}\right )+\frac {1}{4} x^{3/2} (a-b x)^{5/2}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {5}{8} a \left (\frac {1}{2} a \left (\frac {1}{4} a \left (\frac {a \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a-b x}}\right )}{b^{3/2}}-\frac {\sqrt {x} \sqrt {a-b x}}{b}\right )+\frac {1}{2} x^{3/2} \sqrt {a-b x}\right )+\frac {1}{3} x^{3/2} (a-b x)^{3/2}\right )+\frac {1}{4} x^{3/2} (a-b x)^{5/2}\) |
(x^(3/2)*(a - b*x)^(5/2))/4 + (5*a*((x^(3/2)*(a - b*x)^(3/2))/3 + (a*((x^( 3/2)*Sqrt[a - b*x])/2 + (a*(-((Sqrt[x]*Sqrt[a - b*x])/b) + (a*ArcTan[(Sqrt [b]*Sqrt[x])/Sqrt[a - b*x]])/b^(3/2)))/4))/2))/8
3.6.53.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[2 Sub st[Int[1/(b - d*x^2), x], x, Sqrt[b*x]/Sqrt[c + d*x]], x] /; FreeQ[{b, c, d }, x] && !GtQ[c, 0]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Time = 0.08 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.84
method | result | size |
risch | \(-\frac {\left (-48 b^{3} x^{3}+136 a \,b^{2} x^{2}-118 a^{2} b x +15 a^{3}\right ) \sqrt {x}\, \sqrt {-b x +a}}{192 b}+\frac {5 a^{4} \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {a}{2 b}\right )}{\sqrt {-b \,x^{2}+a x}}\right ) \sqrt {x \left (-b x +a \right )}}{128 b^{\frac {3}{2}} \sqrt {x}\, \sqrt {-b x +a}}\) | \(102\) |
default | \(\frac {x^{\frac {3}{2}} \left (-b x +a \right )^{\frac {5}{2}}}{4}+\frac {5 a \left (\frac {x^{\frac {3}{2}} \left (-b x +a \right )^{\frac {3}{2}}}{3}+\frac {a \left (\frac {x^{\frac {3}{2}} \sqrt {-b x +a}}{2}+\frac {a \left (-\frac {\sqrt {x}\, \sqrt {-b x +a}}{b}+\frac {a \sqrt {x \left (-b x +a \right )}\, \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {a}{2 b}\right )}{\sqrt {-b \,x^{2}+a x}}\right )}{2 b^{\frac {3}{2}} \sqrt {x}\, \sqrt {-b x +a}}\right )}{4}\right )}{2}\right )}{8}\) | \(121\) |
-1/192*(-48*b^3*x^3+136*a*b^2*x^2-118*a^2*b*x+15*a^3)/b*x^(1/2)*(-b*x+a)^( 1/2)+5/128/b^(3/2)*a^4*arctan(b^(1/2)*(x-1/2*a/b)/(-b*x^2+a*x)^(1/2))*(x*( -b*x+a))^(1/2)/x^(1/2)/(-b*x+a)^(1/2)
Time = 0.23 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.36 \[ \int \sqrt {x} (a-b x)^{5/2} \, dx=\left [-\frac {15 \, a^{4} \sqrt {-b} \log \left (-2 \, b x + 2 \, \sqrt {-b x + a} \sqrt {-b} \sqrt {x} + a\right ) - 2 \, {\left (48 \, b^{4} x^{3} - 136 \, a b^{3} x^{2} + 118 \, a^{2} b^{2} x - 15 \, a^{3} b\right )} \sqrt {-b x + a} \sqrt {x}}{384 \, b^{2}}, -\frac {15 \, a^{4} \sqrt {b} \arctan \left (\frac {\sqrt {-b x + a}}{\sqrt {b} \sqrt {x}}\right ) - {\left (48 \, b^{4} x^{3} - 136 \, a b^{3} x^{2} + 118 \, a^{2} b^{2} x - 15 \, a^{3} b\right )} \sqrt {-b x + a} \sqrt {x}}{192 \, b^{2}}\right ] \]
[-1/384*(15*a^4*sqrt(-b)*log(-2*b*x + 2*sqrt(-b*x + a)*sqrt(-b)*sqrt(x) + a) - 2*(48*b^4*x^3 - 136*a*b^3*x^2 + 118*a^2*b^2*x - 15*a^3*b)*sqrt(-b*x + a)*sqrt(x))/b^2, -1/192*(15*a^4*sqrt(b)*arctan(sqrt(-b*x + a)/(sqrt(b)*sq rt(x))) - (48*b^4*x^3 - 136*a*b^3*x^2 + 118*a^2*b^2*x - 15*a^3*b)*sqrt(-b* x + a)*sqrt(x))/b^2]
Result contains complex when optimal does not.
Time = 11.40 (sec) , antiderivative size = 326, normalized size of antiderivative = 2.69 \[ \int \sqrt {x} (a-b x)^{5/2} \, dx=\begin {cases} \frac {5 i a^{\frac {7}{2}} \sqrt {x}}{64 b \sqrt {-1 + \frac {b x}{a}}} - \frac {133 i a^{\frac {5}{2}} x^{\frac {3}{2}}}{192 \sqrt {-1 + \frac {b x}{a}}} + \frac {127 i a^{\frac {3}{2}} b x^{\frac {5}{2}}}{96 \sqrt {-1 + \frac {b x}{a}}} - \frac {23 i \sqrt {a} b^{2} x^{\frac {7}{2}}}{24 \sqrt {-1 + \frac {b x}{a}}} - \frac {5 i a^{4} \operatorname {acosh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{64 b^{\frac {3}{2}}} + \frac {i b^{3} x^{\frac {9}{2}}}{4 \sqrt {a} \sqrt {-1 + \frac {b x}{a}}} & \text {for}\: \left |{\frac {b x}{a}}\right | > 1 \\- \frac {5 a^{\frac {7}{2}} \sqrt {x}}{64 b \sqrt {1 - \frac {b x}{a}}} + \frac {133 a^{\frac {5}{2}} x^{\frac {3}{2}}}{192 \sqrt {1 - \frac {b x}{a}}} - \frac {127 a^{\frac {3}{2}} b x^{\frac {5}{2}}}{96 \sqrt {1 - \frac {b x}{a}}} + \frac {23 \sqrt {a} b^{2} x^{\frac {7}{2}}}{24 \sqrt {1 - \frac {b x}{a}}} + \frac {5 a^{4} \operatorname {asin}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{64 b^{\frac {3}{2}}} - \frac {b^{3} x^{\frac {9}{2}}}{4 \sqrt {a} \sqrt {1 - \frac {b x}{a}}} & \text {otherwise} \end {cases} \]
Piecewise((5*I*a**(7/2)*sqrt(x)/(64*b*sqrt(-1 + b*x/a)) - 133*I*a**(5/2)*x **(3/2)/(192*sqrt(-1 + b*x/a)) + 127*I*a**(3/2)*b*x**(5/2)/(96*sqrt(-1 + b *x/a)) - 23*I*sqrt(a)*b**2*x**(7/2)/(24*sqrt(-1 + b*x/a)) - 5*I*a**4*acosh (sqrt(b)*sqrt(x)/sqrt(a))/(64*b**(3/2)) + I*b**3*x**(9/2)/(4*sqrt(a)*sqrt( -1 + b*x/a)), Abs(b*x/a) > 1), (-5*a**(7/2)*sqrt(x)/(64*b*sqrt(1 - b*x/a)) + 133*a**(5/2)*x**(3/2)/(192*sqrt(1 - b*x/a)) - 127*a**(3/2)*b*x**(5/2)/( 96*sqrt(1 - b*x/a)) + 23*sqrt(a)*b**2*x**(7/2)/(24*sqrt(1 - b*x/a)) + 5*a* *4*asin(sqrt(b)*sqrt(x)/sqrt(a))/(64*b**(3/2)) - b**3*x**(9/2)/(4*sqrt(a)* sqrt(1 - b*x/a)), True))
Time = 0.29 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.39 \[ \int \sqrt {x} (a-b x)^{5/2} \, dx=-\frac {5 \, a^{4} \arctan \left (\frac {\sqrt {-b x + a}}{\sqrt {b} \sqrt {x}}\right )}{64 \, b^{\frac {3}{2}}} + \frac {\frac {15 \, \sqrt {-b x + a} a^{4} b^{3}}{\sqrt {x}} + \frac {55 \, {\left (-b x + a\right )}^{\frac {3}{2}} a^{4} b^{2}}{x^{\frac {3}{2}}} + \frac {73 \, {\left (-b x + a\right )}^{\frac {5}{2}} a^{4} b}{x^{\frac {5}{2}}} - \frac {15 \, {\left (-b x + a\right )}^{\frac {7}{2}} a^{4}}{x^{\frac {7}{2}}}}{192 \, {\left (b^{5} - \frac {4 \, {\left (b x - a\right )} b^{4}}{x} + \frac {6 \, {\left (b x - a\right )}^{2} b^{3}}{x^{2}} - \frac {4 \, {\left (b x - a\right )}^{3} b^{2}}{x^{3}} + \frac {{\left (b x - a\right )}^{4} b}{x^{4}}\right )}} \]
-5/64*a^4*arctan(sqrt(-b*x + a)/(sqrt(b)*sqrt(x)))/b^(3/2) + 1/192*(15*sqr t(-b*x + a)*a^4*b^3/sqrt(x) + 55*(-b*x + a)^(3/2)*a^4*b^2/x^(3/2) + 73*(-b *x + a)^(5/2)*a^4*b/x^(5/2) - 15*(-b*x + a)^(7/2)*a^4/x^(7/2))/(b^5 - 4*(b *x - a)*b^4/x + 6*(b*x - a)^2*b^3/x^2 - 4*(b*x - a)^3*b^2/x^3 + (b*x - a)^ 4*b/x^4)
Timed out. \[ \int \sqrt {x} (a-b x)^{5/2} \, dx=\text {Timed out} \]
Timed out. \[ \int \sqrt {x} (a-b x)^{5/2} \, dx=\int \sqrt {x}\,{\left (a-b\,x\right )}^{5/2} \,d x \]